Question

An insurance company insured 2000 scooter driver,4000 car drivers and 6000 truck drivers.The probability of accidents are 0.01,0.03 and 0.15 respectively.One of the insured person meets with an accident.What is the probability that he is a scooter driver?

Answer 1

The correct answer is:$\frac{1}{52}$
Let $E_1$=Person chosen is a scooter driver, $E_2$=Person chosen is a car driver, $E_3$=Person chosen is a truck driver and $A$=Person meets with an accident
Since there are 12000 persons, therefore,
Now $P(E_1)=\frac{2000}{12000}=\frac{1}{6}, P(E_2)=\frac{4000}{12000}=\frac{1}{3}, P(E_3)=\frac{6000}{12000}=\frac{1}{2}$
It is given that $P(A|E_1)=$P(a person meets with an accident, he is a scooter driver)=0.01
Similarly, $P(A|E_2)=0.03$ and $P(A|E_3)=0.15$
To find: $P$(person meets with an accident that he is a scooter driver)
Therefore, by Bayes'theorem,
$P(E_1|A)=\frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)+P(E_3)P(A|E_3)}$
$=\frac{\frac{1}{6}×0.001}{\frac{1}{6}×0.001+\frac{1}{3}×0.03+\frac{1}{2}×0.15}$
$=\frac{\frac{1}{6}}{\frac{104}{12}}$
$=\frac{1}{52}$